Answer
$P(X=2)=C(5,2)*0.1^2*0.9^3=\frac{5!}{2!\times(5-2)!}\times0.1^2\times0.9^3=0.0729$
The probability is $0.0729$
Work Step by Step
The solution is based on the binomial distribution:
$P(X=x)=C(n,x)*p^{x}*q^{n-x}$
If we perform 5 independent Bernoulli trials, and $p=0.1$ the probability of two successes are:
$C(5,2)*0.1^2*0.9^3$.
As there are 5 trial out of which two are successes. And the probability of a success is 0.1.
$P(X=2)=C(5,2)*0.1^2*0.9^3=\frac{5!}{2!\times(5-2)!}\times0.1^2\times0.9^3=0.0729$
